question_answer

Let R = radius of a spherical refracting surface and l= least distance between conjugate focii. Then, refractive index of medium of refracting surface is

A) \[\frac{R+l}{R-l}\]

B) \[{{\left( \frac{R+l}{R-l} \right)}^{2}}\]

C) \[\frac{2\,(R-l)}{R+l}\]

D) \[2{{\left( \frac{R-l}{R+1} \right)}^{2}}\]

Correct Answer: B

Solution :

For a spherical refracting surface,

\[\frac{\mu }{v}-\frac{1}{\left( -u \right)}=\frac{\mu -1}{R}\]\[\Rightarrow \frac{\mu }{v}+\frac{1}{\left( u \right)}=\frac{\mu -1}{R}\]

Let \[u+v=x\]

\[\Rightarrow \,\,\,\frac{\mu }{x-u}+\frac{1}{u}=\frac{\mu -1}{R}\]\[\Rightarrow \,\,\,x=u+\frac{\mu Ru}{\left( \mu -1 \right)u-R}\]

\[\therefore \frac{du}{dx}=\frac{\mu Ru}{{{\left( \left( \mu -1 \right)u-R \right)}^{2}}}\]

L is minimum, when \[\frac{du}{dx=0}\]

\[\therefore u=\frac{R}{\sqrt{\left( \mu  \right)}-1}\]

From Eq. (i), we have

\[v=\frac{\sqrt{\mu }\,\,R}{\sqrt{\mu }-1}\]

As \[l=u+v,l=\frac{R}{\sqrt{\mu }-1}+\frac{\sqrt{\mu }.R}{\sqrt{\mu }-1}\]

\[\mu ={{\left( \frac{R+l}{R-l} \right)}^{2}}\]